EE247 Lecture 4. For simplicity, will start with all pole ladder type filters. Convert to integrator based form- example shown

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1 EE247 Lecture 4 Ldder type filters For simplicity, will strt with ll pole ldder type filters Convert to integrtor bsed form exmple shown Then will ttend to high order ldder type filters incorporting zeros Implement the sme 7th order elliptic filter in the form of ldder C with zeros Find level of sensitivity to component mismtch Compre with cscde of biquds Convert to integrtor bsed form utilizing SFG techniques Effect of Integrtor NonIdelities on Filter Frequency Chrcteristics EECS 247 Lecture 4: Active Filters 2007 H.K. Pge LC Ldder Filters L2 L4 in C C3 C5 Design: Filter tbles A. Zverev, Hndbook of filter synthesis, Wiley, 967. A. B. Willims nd F. J. Tylor, Electronic filter design, 3 rd edition, McGrwHill, 995. CAD tools Mtlb Spice EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 2

2 LC Ldder Filter Design Exmple Design LPF with mximlly flt pssbnd: f3db = 0MHz, fstop = 20MHz >27dB Mximlly flt pssbnd Butterworth Find minimum filter order : Use of Mtlb or Tbles Here tbles used fstop / f3db = 2 >27dB 3dB Stopbnd Attenution db Minimum Filter Order 5th order Butterworth 2 Νοrmlized ω From: Willims nd Tylor, p. 237 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 3 LC Ldder Filter Design Exmple Find vlues for L & C from Tble: Note L &C vlues normlized to ω 3dB = Denormliztion: Multiply ll L Norm, C Norm by: L r = R/ω 3dB C r = /(RXω 3dB ) R is the vlue of the source nd termintion resistor (choose both Ω for now) Then: L= L r xl Norm C= C r xc Norm From: Willims nd Tylor, p..3 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 4

3 LC Ldder Filter Design Exmple Find vlues for L & C from Tble: Normlized vlues: C Norm =C5 Norm =0.68 C3 Norm = 2.0 L2 Norm = L4 Norm =.68 Denormliztion: Since ω 3dB =2πx0MHz L r = R/ω 3dB = 5.9 nh C r = /(RXω 3dB )= 5.9 nf R = C=C5=9.836nF, C3=3.83nF L2=L4=25.75nH From: Willims nd Tylor, p..3 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 5 Lst Lecture: Exmple: 5 th Order Butterworth Filter in =Ohm L2=25.75nH C 9.836nF L4=25.75nH C3 3.83nF C nF =Ohm Specifictions: f3db = 0MHz, fstop = 20MHz >27dB Used filter tbles to obtin Ls & Cs Mgnitude (db) SPICE simultion Results 30dB Frequency [MHz] EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 6

4 LowPss C Ldder Filter Conversion to Integrtor Bsed Active Filter in I L2 I3 L4 I 5 C C3 C5 I 2 I4 I 6 I 7 Use KCL & KL to derive equtions: I = in 2, 2 2 =, 3 = 2 sc 4 I I =, 5 = 4 6, 6 = o = 6 sc 3 sc 5 3 I =, I 2 = I I 3, I 3 = sl2 5 6 I 4 = I 3 I 5, I 5 =, I 6 = I 5 I 7, I7 = sl4 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 7 I LowPss C Ldder Filter Signl Flowgrph = in 2, I 2 2 = sc, 3 = 2 4 I I =, 5 = 4 6, 6 = sc 3 sc 5 o = 6 3 I =, I 2 = I I 3, I3 = sl2 5 6 I 4 = I 3 I 5, I 5 =, I 6 = I 5 I 7, I7 = sl4 in 2 sc sl2 sc3 sl4 sc5 I o I3 I4 I 5 I 6 I 7 SFG EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 8

5 LowPss C Ldder Filter Signl Flowgrph in I L2 I3 L4 I 5 C C3 C5 I 2 I4 I 6 I 7 in 2 sc sl2 sc3 sl4 sc5 o I I 2 I3 I4 I 5 I 6 I 7 SFG EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 9 in 2 sc I LowPss C Ldder Filter Normlize I sl2 sc3 sl4 sc5 o I 3 I4 I 5 I 6 I 7 in * R * * R R sc * R sc * sl 3R sl sc * 2 4 5R o * R 7 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 0

6 in * R LowPss C Ldder Filter Synthesize * * R R sc * R sc * sl 3R sl sc * 2 4 5R o * R 7 in * R sτ 2 sτ sτ 3 sτ 4 sτ 5 * R 3 5 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge LowPss C Ldder Filter Integrtor Bsed Implementtion in * R sτ 2 sτ sτ 3 sτ 4 sτ 5 * R 3 * L2 * * L4 * * = C.R, 2 = = C.R, C.R, C.R, C.R * 2 3 = 3 4 = = * 4 5 = 5 R R τ τ τ τ τ 5 Building Block: RC Integrtor 2 = src EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 2

7 Negtive Resistors o 2 o 2 o 2 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 3 Integrtor Bsed Implementtion of LP Ldder Filter Synthesize EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 4

8 Frequency Response MHz 0MHz EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 5 Scle Node oltges Scle by fctor s EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 6

9 Node Scling 2 3 X.2 X.2 X.2/.6 X /.2 X.6/.2 4 X.6 5 X.8 X.8/2 X.8/.6 X.6/.8 X 2/.8 O X 2 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 7 Mximizing Signl Hndling by Node oltge Scling Before Node Scling After Node Scling Scle by fctor s EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 8

10 Filter Noise Totl the output:.4 μ rms (noiseless opmps) Tht s excellent, but the cpcitors re very lrge (nd the resistors smll high power dissiption). Not possible to integrte. Suppose our ppliction llows higher noise in the order of 40 μ rms EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 9 Scle to Meet Noise Trget Scle cpcitors nd resistors to meet noise objective s = 0 4 Noise: 4 μ rms (noiseless opmps) EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 20

11 Completed Design th order ldder filter Finl design utilizing: Node scling Finl R & C scling bsed on noise considertions 5 4 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 2 Sensitivity C mde (rbitrrily) 50% (!) lrger thn its nominl vlue 0.5 db error t bnd edge 3.5 db error in stopbnd Looks like very low sensitivity EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 22

12 Differentil 5 th Order Lowpss Filter in Since ech signl nd its inverse redily vilble, elimintes the need for negtive resistors! Differentil design hs the dvntge of even order hrmonic distortion components nd common mode spurious pickup utomticlly cncels Disdvntge: Double resistor nd cpcitor re! EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 23 C Ldder Filters Including Trnsmission Zeros All poles in C L2 C3 L4 C5 Poles & Zeros C2 C4 C6 in C L2 C3 L4 C5 L6 C7 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 24

13 C Ldder Filter Design Exmple Design bsebnd filter for CDMA IS95 cellulr phone receive pth with the following specs. Filter frequency msk shown on the next pge Allow enough mrgin for mnufcturing vritions Assume pssbnd mgnitude vrition of.8db Assume the 3dB frequency cn vry by 8% due to mnufcturing tolernces & circuit inccurcies Assume ny phse impirment cn be compensted in the digitl domin * Note this is the sme exmple s for cscde of biqud while the specifictions re given closer to rel product cse EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 25 C Ldder Filter Design Exmple CDMA IS95 Receive Filter Frequency Msk 0 Mgnitude (db) k 700k 900k.2M Frequency [Hz] EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 26

14 C Ldder Filter Design Exmple: CDMA IS95 Receive Filter Since phse impirment cn be corrected for, use filter type with mx. rolloff slope/pole Filter type Elliptic Design filter freq. response to fll well within the freq. msk Allow mrgin for component vritions & mismtches For the pssbnd ripple, llow enough mrgin for ripple chnge due to component & temperture vritions Design nominl pssbnd ripple of 0.2dB For stopbnd rejection dd few db mrgin 445=49dB Finl design specifictions: fpss = 650 khz Rpss = 0.2 db fstop = 750 khz top = 49 db Use Mtlb or filter tbles to decide the min. order for the filter (sme s cscded biqud exmple) 7 th Order Elliptic EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 27 C LowPss Ldder Filter Design Exmple: CDMA IS95 Receive Filter 7 th order Elliptic C2 C4 C6 in C L2 C3 L4 C5 L6 C7 Use filter tbles to determine LC vlues EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 28

15 C Ldder Filter Design Exmple: CDMA IS95 Receive Filter Specifictions fpss = 650 khz Rpss = 0.2 db fstop = 750 khz top = 49 db Use filter tbles to determine LC vlues Tble from: A. Zverev, Hndbook of filter synthesis, Wiley, 967 Elliptic filters tbulted wrt reflection coeficient ρ Rpss= 0 log( ρ 2 ) Since Rpss=0.2dB ρ =20% Use tble ccordingly EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 29 C Ldder Filter Design Exmple: CDMA IS95 Receive Filter Tble from Zverev book pge #28 & 282: Since our spec. is Amin=44dB dd 5dB mrgin & design for Amin=49dB EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 30

16 Tble from Zverev pge #28 & 282: Normlized component vlues: C=.7677 C2= L2=.9467 C3=.534 C4=.0098 L4= C5= C6=0.72 L6= C7= EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 3 C Filter Frequency Response Frequency msk superimposed Frequency response well within spec. Mgnitude (db) Frequency [khz] EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 32

17 Pssbnd Detil Pssbnd well within spec Mgnitude (db) Frequency [khz] EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 33 C Ldder Filter Sensitivity The design hs the sme specifictions s the previous exmple implemented with cscded biquds To compre the sensitivity of C ldder versus cscdedbiquds: Chnged ll Ls &Cs one by one by 2% in order to chnge the pole/zeros by % (similr test s for cscded biqud) Found frequency response most sensitive to L4 vritions Note tht by vrying L4 both poles & zeros re vried EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 34

18 RCL Ldder Filter Sensitivity Component mismtch in C filter: Increse L4 from its nominl vlue by 2% Decrese L4 by 2% Mgnitude (db) L4 nom L4 low L4 high Frequency [khz] EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 35 RCL Ldder Filter Sensitivity 5 0.2dB 5 Mgnitude (db) dB Frequency [khz] EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 36

19 Sensitivity of Cscde of Biquds Component mismtch in Biqud 4 (highest Q pole): Increse ω p4 by % Decrese ω z4 by % 2.2dB Mgnitude (db) dB kHz 600kHz Frequency [Hz] MHz High Q poles High sensitivity in Biqud reliztions EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 37 Sensitivity Comprison for CscdedBiquds versus C Ldder 7 th Order elliptic filter % chnge in pole & zero pir Pssbnd devition Stopbnd devition Cscded Biqud 2.2dB (29%) 3dB (40%) C Ldder 0.2dB (2%).7dB (2%) Doubly terminted LC ldder filters Significntly lower sensitivity compred to cscdedbiquds prticulrly within the pssbnd EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 38

20 7 th order Elliptic C Ldder Filter Design Exmple: CDMA IS95 Receive Filter C2 C4 C6 in C L2 C3 L4 C5 L6 C7 Previously lerned to design integrtor bsed ldder filters without trnsmission zeros Question: o How do we implement the trnsmission zeros in the integrtorbsed version? o Preferred method no extr power dissiption no ctive elements EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 39 Integrtor Bsed Ldder Filters How Do to Implement Trnsmission zeros? C in I 2 3 L2 I 3 C I 2 I 4 4 C3 I 5 Use KCL & KL to derive : I I I3 IC I 2 2 = I I3 I C, IC = ( 2 4) s C, 2 =, sc 2 = sc Substituting for IC nd rerrnging : I I3 C 2 = s 4 ( C C ) C C EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 40

21 Integrtor Bsed Ldder Filters How Do to Implement Trnsmission zeros? C in I 2 3 L2 I 3 C I 2 I 4 4 C3 I 5 Use KCL & KL to derive : I I C 3 2 = 4 s( C C ) C C I3 I5 C 4 = s 2 ( C 3 C ) C 3 C Frequency independent constnts Cn be substituted by: oltgecontrolled oltge Source EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 4 Integrtor Bsed Ldder Filters Trnsmission zeros C in I 2 3 L2 I 3 C C I 2 4 I 5 ( ) ( C 3 C ) C 4 C C C I 4 2 C 3 C Replce shunt cpcitors with voltge controlled voltge sources: I I3 C 2 = s 4 ( C C ) C C I3 I5 C 4 = s 2 ( C 3 C ) C 3 C EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 42

22 3 rd Order Lowpss Filter All Poles & No Zeros in I 2 3 L2 I 3 C I2 I 4 4 C 3 I 5 in 2 sc 3 4 sl2 sc 3 o I I 2 I 3 I 4 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 43 in Trnsmission Zero Implementtion W/O Use of Active Elements I I 5 L2 I 3 ( C C ) ( C 3 C ) C C 4 C C I 2 I 2 4 C 3 C C C C C C 3 C in s C C sl2 s( C3 C) ( ) I I 2 I 3 I 4 o EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 44

23 Integrtor Bsed Ldder Filters Higher Order Trnsmission zeros 2 C 4 C b 6 Convert zero generting Cs in C loops to voltgecontrolled voltge sources C C ( ) ( ) C C C 3 C C b C C 4 C C 2 C C 3 C b 6 C 3 C b C 5 6 ( C 5 C b) C b 4 C 3 C b EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 45 in in 2 * R s( C C ) R * Higher Order Trnsmission zeros I I 7 I L2 I 3 I L4 4 ( C C ) ( C 3 C C b) ( C 5 C b) C C 2 C 4 C C C C I 3 b 4 C C 2 I 6 5 C b b 6 C 3 C b C C C b C C C C 3 C C 3 C b b C 5 C b * R * R sr * sl2 ( C3 C Cb) sl4 sr * C C ( 5 b ) EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 46 o * R 7

24 Exmple: 5 th Order Chebyshev II Filter 5 th order Chebyshev II Tble from: Willims & Tylor book, p..2 50dB stopbnd ttenution f 3dB =0MHz EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 47 Reliztion with Integrtor C = sc C R i 2 3 ( ) * C C R* EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 48

25 5 th Order Butterworth Filter 2 From: Lecture 4 pge EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 49 5 th Order Chebyshev II Filter OpmpRC Simultion EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 50

26 in 7th Order Differentil Lowpss Filter Including Trnsmission Zeros Trnsmission zeros implemented with pir of coupling cpcitors EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 5 Effect of Integrtor NonIdelities on Filter Frequency Chrcteristics In the pssive filter design (C filters) section: Rective element (L & C) nonidelities expressed in the form of Qulity Fctor (Q) Filter impirments due to component nonidelities explined in terms of component Q In the context of ctive filter design (integrtorbsed filters) Integrtor nonidelities Trnslted to hve form of Qulity Fctor (Q) Filter impirments due to integrtor nonidelities explined in terms of integrtor Q EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 52

27 Effect of Integrtor NonIdelities on Filter Performnce Idel integrtor chrcteristics Rel integrtor chrcteristics: Effect of opmp finite DC gin Effect of integrtor nondominnt poles EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 53 Effect of Integrtor NonIdelities on Filter Performnce Idel Integrtor C R in Idel opmp DC gin= Single DC no nondominnt poles ω H(s) = o s ωo = / RC 0dB ψ 90 o Idel Idel Intg. Intg. log H ( s) ω 0 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 54

28 Idel Integrtor Qulity Fctor Idel intg. trnsfer function: Since component Q is defined s:: ωo ωo H(s) = = = s jω ω j ωo H Q ( jω ) = R( ω) jx( ω) X ( ω) = R ( ω) Then: intg. Q idel = EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 55 Rel Integrtor NonIdelities Idel Intg. log H ( s) Rel Intg. log H ( s) ω 0 ω P = 0 ω 0 P2P3 ψ ψ 90 o 90 o ωo H(s) = H(s) s s s ( s o )( p2)( ω p3)... EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 56

29 Effect of Integrtor Finite DC Gin on Q log H ( s) ψ ω P = 0 ω ω o ω π ArctnP 2 ωo P Phse ω ω o o (in rdin) 90 o Exmple: P/ ω 0 = /00 phse error 0.5degree EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 57 Effect of Integrtor Finite DC Gin on Q Idel intg Intg with finite DC gin Phse ω 0 Droop in the pssbnd Mgnitude (db) Droop in the pssbnd Normlized Frequency EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 58

30 Effect of Integrtor NonDominnt Poles log H ( s) ψ ω 0 P2P ω o ω ω π Arctn o 2 p i= 2 i ωo Phse ω p o i= 2 i (in rdin) 90 o Exmple: ω 0 /P2 =/00 phse error 0.5degree EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 59 Effect of Integrtor NonDominnt Poles Idel intg Opmp with finite bndwidth Mgnitude (db) Peking in the pssbnd Phse ω 0 Peking in the pssbnd In extreme cses could result in oscilltion! Normlized Frequency EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 60

31 Effect of Integrtor NonDominnt Poles & Finite DC Gin on Q log H ( s ) ω P = 0 ψ ω 0 P2P3 ω o ω π Arctn 2 P ωo ω Arctn o p 90 i = 2 i 90 o Note tht the two terms hve different signs Cn cncel ech other s effect! EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 6 Integrtor Qulity Fctor Rel intg. trnsfer function: Bsed on the definition of Q nd ssuming tht: H(s) s... ( s s )( p2)( ω p3) o ωo << & >> p 2,3,... It cn be shown tht in the vicinity of unityginfrequency: Q intg. rel ω o i= 2 p i Phse ω 0 Phse ω 0 EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 62

32 Exmple: Effect of Integrtor Finite Q on Bndpss Filter Behvior 0.5 ο φ ω o intg 0.5 ο φ ω o intg Idel Idel Integrtor DC gin=00 Integrtor 00.ω o EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 63 Exmple: Effect of Integrtor Q on Filter Behvior ( 0.5 ο φ led 0.5 ο φ excess ω o intg φ ω o intg ~ 0 Idel Integrtor DC gin=00 & 00. ω ο EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 64

33 Summry Effect of Integrtor NonIdelities on Q Q intg. idel = Q intg. rel ω o p i= 2 i Amplifier DC gin reduces the overll Q in the sme mnner s series/prllel resistnce ssocited with pssive elements Amplifier poles locted bove integrtor unitygin frequency enhnce the Q! If nondominnt poles close to unitygin freq. Oscilltion Depending on the loction of unityginfrequency, the two terms cn cncel ech other out! EECS 247 Lecture 4: Active Filters 2007 H.K. Pge 65

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